Please help me!!!!!!!!!!!!!!!!!! We are desperate!
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The perimeter of a geometric shape is the sum of the lengths of all sides of the shape.
Since the triangle is the geometric shape that has three sides, the perimeter is the sum of the lengths of these sides.
Put the length of the sides of the triangle: x,y,z.
P = x+y+z.
Since the triangle is right, then one side is known as the hypotenuse of triangle and the other two sides are known as catheti of triangle.
The Pythagora's theorem acts in a right triangle such that: the square of the length of hypotenuse is equal to the sum of the squares of the lengths of catheti.
Put x = hypotenuse and y,z = catheti.
x^2 = y^2 + z^2
The length of one cathetus is three times longer than the other cathetus => y = 3z
Since the perimeter is of 12m => 12 = x + 3z + z => 12 = x + 4z => x = 12 - 4z
Write for z the Pythagora's theorem:
(12 - 4z)^2 = (3z)^2 + z^2
144 - 96z + 16z^2 = 9z^2 + z^2
Subtract 10z^2 both sides:
144 - 96z + 16z^2 - 10z^2 = 0
6z^2 - 96z + 144 = 0
Divide the quadratic by 6:
z^2 - 16z + 24 = 0
Use quadratic formula: z_(1,2) = (16+-sqrt(256 - 96 ))/2
z_(1,2) = (16+-4sqrt(10))/2 => z_(1,2) = (8+-2sqrt(10))
z_(1,2) = 8+-6.324
z_1 = 14.324 and z_2 = 1.676
Since the perimeter of the right triangle is 12 m, the lengths of all sides must be less than this value => the length of the cathetus z cannot be of 14.32 m, therefore you need to consider only the value z_2 = 1.676 m.
The other cathetus is three times z => y = 5.028 m
x = 12 - (1.676 + 5.028) => x = 5.296 m
The lengths of the sides of the right triangle are: x = 5.296 m ; y = 5.028 m ; z_2 = 1.676 m.
Right triangle dimensions: a,b and c, where c is hypotenuse.
b = 3a
a + b + c= 12
a + 3a + c = 12
4a + c = 12
Now, Pythagorean Law is a^2 + b^2 = c^2
a^ +(3a)^2 = c^2
10a^2 = c^2
c = Sq. Rt. 10 (a) = 3.2a
4a + 3.2a = 12
7.2a = 12
a = 1.7
b = 3a = 3(1.7) = 5.1
c^2 = a^2 + b^2
c^2 = 1.7^2 + 5.1^2 = 2.9 + 26 = 28.9
c = sq. root 28.9 = 5.4
So sides are a = 1.7, b = 5.1 and c = 5.4
I ment "Find each side of the triangle to the nearest tenth." Sorry
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